Determine KçandKpat 298 K for the reaction: NOBR(g) = % N2{g) + % O2(g) + % Br2(g) from the following information at 298 K: NO(g) ) = % N2(g) + % O2ig) Ke = 1.55 x 1015 ½ NO(g) + % Br2(g) = % NOBr(a) Ko = = 2.63 Type your answers below. Note: If answering in scientific notation, type your answer this way 1.23 x 1045 → type 1.23e45 ANSWERS: Kc = Kp =

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter12: Gaseous Chemical Equilibrium
Section: Chapter Questions
Problem 18QAP: Given the following data at 25C 2NO(g)N2(g)+O2(g)K=1 10 30 2NO(g)+Br2(g)2NOBr(g)K=8 101 Calculate K...
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Determine Kandk,at 298 K for the reaction:
NOBR(g) = % N2(g) + O2(g) + % Br2(g)
from the following information at 298 K:
NO(9) = % N2(g) + % O2(a)
Ke = 1.55 x 1015
% NO(g) + % Br2(g) = % NOBr(a)
Kp = 2.63
Type your answers below.
Note: If answering in scientific notation, type your answer this way:
1.23 x 1045 → type 1.23e45
ANSWERS:
Kc =
Kp =
Transcribed Image Text:Determine Kandk,at 298 K for the reaction: NOBR(g) = % N2(g) + O2(g) + % Br2(g) from the following information at 298 K: NO(9) = % N2(g) + % O2(a) Ke = 1.55 x 1015 % NO(g) + % Br2(g) = % NOBr(a) Kp = 2.63 Type your answers below. Note: If answering in scientific notation, type your answer this way: 1.23 x 1045 → type 1.23e45 ANSWERS: Kc = Kp =
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